Karoubi envelope of a ring as a category Given a (not necessarily commutative) ring $R$. We can consider $R$ as an additive category $\mathcal R$ with one object $*$ and homomorphisms $\hom(*,*)=R$. One can form the idempotent completion of $\mathcal R$ (aka its Karoubi envelope), which is the universal category where every idempotent of $\mathcal R$ splits.
Question: What is the idempotent completion in this special case? Is there any ring-theoretic description of it? Is it, for instance, again a ring?
 A: For a small category $C$, the idempotent completion can be identified with the full subcategory of presheaves on $C$ which are retracts of objects in $C$. 
In your case, a functor $F:R\to Set$ is simply a set $S:=F(\ast)$ equipped with a left $R$-action. The yoneda embedding of the unique object $\ast\in R$ is simply $R$, viewed as a set equipped with the obvious $R$-action. If $S$ is a retract of $R$, then there are morphisms of lets $R$-sets $i:S\to R$ and $r:R\to S$ such that $ri={\rm id}:S\to S$. 
Since $i$ and $r$ are in particular functions, $i$ must be injective. Since $i$ is a map of left $R$-sets, the image of $i$ must be closed under the $R$-action. Moreover, $i(S)$ must actually be a principal ideal in $R$, since $r:R\to S$ is surjective and therefore $i(r(1))\in i(S)$ must generate $\ i(S)$. Even more, $i(r(1))$ must be idempotent, since $rir=r$. 
So the category you are looking for is the category of principal ideals, generated by idempotents in $R$, with morphisms given by $R$-module maps between them. 
Notice that if you have an idempotent $e\in R$, then the corresponding map in the idempotent completion is multiplication on the left by $e$: 
$$e:R\to R$$
This idempotent is split since we can take $S=eR$ and then $e=i\circ e$ and ${\rm id}=e\circ i$.
